I know this is a humor subreddit and this is a joke, but this problem wasted a huge week of mine since I was dealing with absurdly small numbers in my simulations. Use fsum from math library in Python to solve this people.
Yeah, this works especially well for currencies (effectively doing all calculations in cents/pennies), as you do need perfect precision throughout the calculations, but the final results gets rounded to two-digit-precision anyways.
quite a horrible hack, most modern languages have decimal type that handles floating rounding. And if not, you should just use rounding functions to two digits with currency.
The lesson here is that floating point numbers are not exact and that you should never do a straight comparison with them. Instead check to see if they are within some small tolerance of each other. In python that is done with math.isclose(0.1 + 0.2, 0.3).
I know this is a humor subreddit and this is a joke, but this problem wasted a huge week of mine since I was dealing with absurdly small numbers in my simulations. Use fsum from math library in Python to solve this people.
One of my lecturers mentioned a way they would get around this was to store all values as ints and then append a . two character before the final one.
Yeah, this works especially well for currencies (effectively doing all calculations in cents/pennies), as you do need perfect precision throughout the calculations, but the final results gets rounded to two-digit-precision anyways.
quite a horrible hack, most modern languages have decimal type that handles floating rounding. And if not, you should just use rounding functions to two digits with currency.
Fixed point notation. Before floats were invented, that was the standard way of doing it. You needed to keep your equation within certain boundaries.
The lesson here is that floating point numbers are not exact and that you should never do a straight comparison with them. Instead check to see if they are within some small tolerance of each other. In python that is done with
math.isclose(0.1 + 0.2, 0.3).